Solving High School Math Questions About Circles and Straight Lines to Process 30

Organize the equations for this straight line.

y=x+1 - This is a straight line diagonally 45 to the upper right, and the coordinates of the intersection point with the y-axis are (0,1).

Make a straight line perpendicular to it through the origin, y=-x

This line intersects the previous line at h

It can be solved that h(, let m(a,b) be the intersection point of the line y=x+1 and the circle o.

Draw MN parallel to the y-axis and hn parallel to the x-axis.

As can be seen from the figure, hm = root 6

hn = mn = root 2hm = root 12 = root 3

m(a,b)a=root=

b = root 3 + mo 2 = a 2 + b 2 =

mo=The equation for a circle with o as the center of the circle and the radius r is.

x^2+y^2=r^2

where r=mo=

I don't know why such a strange number is given.

In order to get the shortest tangent of the first quadrant, the tangent point should be on the diameter of the circle o45 degrees, and its coordinates are (c,d) and c=d=radius. The tangent is at the intersection of the y-axis y=2d, and the tangent is at the intersection of the x-axis x=2c

Especially that strange number, please check it carefully. The idea of solving the problem should be correct.

It's true that you can master it well

The original equation is as follows: (x-3) square + (y-4) square = 5 square, and the radius is 5.

That is, mo=5, and because mo=150, on=30.

The center of the circle is the same, and it is enough to change the radius, i.e., the equation is: (x-3) square + (y-4) square = 30 square.

x²+y²-6x-8y=0

The polar equation is -6 cos -8 sin =0|om|=ρ，|om|×|on|=150

on|=150/ρ

150/|on|

Both ON and OM polar angles are

Directly substituted as (150 |on|)²150/|on|)(6cosα+8

sinα)=0

on|is denoted by , ie.

150/ρ)²150/ρ)(6cosα+8sinα)=0

Multiply 150-6 cos -8 sin = 0

i.e. 75-3x-4y=0

Find the equation for a straight line with a chord length of 6 2 that passes through the point p(6,-4) and is circumferentially circled x 2+y 2=20.

Solution: This circle is centered at the origin, and the circle with a radius of 2 5 has a chord length of 6 2

According to the Pythagorean theorem: The distance from the center of the circle to the straight line is 2

This becomes a straight line that passes through the point p(6,-4) and the distance from the origin to this line is 2

Let the equation for the straight line be y+4=k(x-6).

The equation for a straight line is kx-y-6k-4=0

The distance from the origin to this line =|-6k-4|(k 2 + 1) = 2 solution yields: k = -7 17 or k = -1

The linear equation is: (-7 17)x+y+110 17=0 or the linear equation is: -x+y+10=0

Find the equation for the circle with the center of the circle on the line x-y-4-0 and the intersection of two circles x 2+y 2-4x-6-0 and x 2+y 2-4y-6=0.

The intersection of x 2 + y 2 -4x-6 = 0 and x 2 + y -4y-6 = 0 is 4x+6 = 4y+6

y=x，2x^2-4x-6=0

x^2-2x-3=0

x=-1,x=3, so the coordinates of the intersection point are (-1,-1)(3,3), and the center of the circle passing through these two points is on the perpendicular bisector of the line segments connecting these two points, and the equation is y-1=-(x-1).

y=-x+2

The intersection point with x-y-4=0 is.

x=3, y=-1, is the center of the circle, and the distance to (-1, -1) is the radius of 4

So the equation is (x-3) 2+(y+1) 2=16

Draw a diagram with a straight line of y=kx +b

Generation p(6,-4) -4=6k+b

The distance from the center of the circle to the straight line is determined by the Pythagorean theorem.

d=|b|/√(1 +k²)|=√(r²-3√2²)→b²/2=1 +k²②

→k=…,b=…

2): Two circles x=y

Find the intersection point is a(3,3),b(-1,-1), ab midpoint d(1,1) then the perpendicular bisector of ab: y=-x

Then the intersection with x-y-4=0 is the center of the circle.

Center c(2,-2).

r=ac=cb=√10

x-2)²+y+2)²=10

1.(x-2)^2 + y+2)^2=82.The straight line passes through the point m(1 2,-1 2), and the chord is shortest when the slope of the chord is 1.

The slope is -m (m+1)=1 and m=-1 2

Distance from m to the center of the circle: 3 root 2

The radius of the circle is 2 times the root 2

Pythagorean theorem: Shortest length = 2 * root number (8 - 9 2) = root number 14 is right?

The formula for the distance between two points: d= [(x1-x2) 2+(y1-y2) 2] Substituting the straight line y=kx+b yields: d= [(x1-x2) 2+k 2(x1-x2) 2]= [(1+k 2)(x1-x2) 2]= (1+k 2) (x1-x2) 2 x1-x2 positive and negative uncertainty d= (1+k 2)|x1-x2|

y=kx+b

Substituting the circular equation gives us a quadratic equation about x.

You can get x1+x2 and x1x2

And then calculate |x1-x2|

Well, it's actually very simple, it's just that it's a slow 、、、 here

First of all, you know x1 and x2.

Take the chord as the hypotenuse, make a right-angled triangle with two sides parallel to the x-axis and parallel to the y-axis, and k is the tan inclination angle, so the hypotenuse = horizontal right-angled side divided by the cos inclination angle. where the horizontal right-angled edge is equal to |x1-x2|, and the reciprocal of cos is the square of 1+tan under the root number. You can take a look at the square of 1+tan under the root number, it's very simple.

1.Let the equations of two straight lines be ax+by+c=0, ax by c=0, and their intersection points are (x0, y0), then.

ax0+by0+c=0

ax0＋by0＋c=0

So ax0+by0+c+n(ax0 by0 c)=0

Obviously, the straight line ax+by+c+n(ax by c)=0 is the passing point (x0, y0), and every value of n it represents a straight line, so it weighs the linear system equation of the intersection of two straight lines, when n=0 represents the first straight line, but it does not contain the second straight line, pay attention to the test when using.

2.The distance between two points is derived using the formula.

Let the intersection point of the line and the circle be a(x1,y1),b(x2,y2), when the slope of the line does not exist, find the coordinates of the intersection point, and directly find the chord length;

When the slope of the straight line exists, then.

Slope k=(y1-y2) (x1-x2).

ab|=√[(x1-x2)²+y1-y2)²|

[(x1-x2)²+y1-y2)²]

(x1-x2)² x1-x2)²+y1-y2)²]/√(x1-x2)²

(x1-x2)² x1-x2)²+y1-y2)²]/(x1-x2)²

(x1-x2)² 1+(y1-y2)²/(x1-x2)²]

(x1-x2)² 1+k²)

(1+k²)*x1+x2)² 4x1x2]

According to the formula for the distance of a point from a straight line.

d = [absolute value of ax0+by0+c] [arithmetic square root of (a2+b2)].

It can be obtained that the distance from the center of the circle (0,2) to the line y=(3)x is 1, i.e. 2 2 in the above formulation (the numerator and denominator of the distance formula are retained). A right triangle with the center of the circle, one of the intersection points and the midpoint of the chord can be obtained in a circle, and then solved by the Pythagorean theorem.

Refers to the distance from the center of the circle (0,2) to the straight line y=(3)x.

Apply the Pythagorean theorem and the perpendicular diameter theorem. Half of the length of the chord (l 2).The radius of the circle 2, the distance from the center of the circle (0,2) to the line y=(3)x (2 2) is exactly one rt

where (2 2) is the distance from the line l:y= 3x to the center of the circle (0,2) d=|2|/√[(3)²+1²]=2/2.

The equation makes use of the Pythagorean theorem of right triangles, where (2 2) 2 is obtained by sin30 degrees = 1 2 radius.

The circle c is x*2+(y-2)*2=25, that is, the center of the circle is (0,2), the radius is 5 of the circle, through the center of the circle t to do tm perpendicular to the chord ab, from the pythagorean theorem tm = 3, then the center of the circle is (0,2), the circle d with a radius of 3, this circle equation x*2+(y-2)*2=9 is tangent to the straight line ab, because ab passes the point (3,8), let ab linear equation be y-8=k(x-3), connect these two equations, because tangent, so =0, the solution k=3 4, Therefore, the linear AB equation is y-8=3 4(x-3).

The equation for the circle yields x 2+(y-2) 2=25, the center of the circle (0,2), and the radius r=5.

Due to |ab|=8<2r, so there are two such straight lines.

From the Pythagorean theorem, the distance d from the center of the circle to the straight line satisfies: d 2+(|ab|2) 2=r 2, therefore, d 2=9 , let the linear l equation be a(x-3)+b(y-8)=0, (the advantage of this setting is to avoid omissions when the slope does not exist).

Then d 2=(-3a-6b) 2 (a 2+b 2)=9, simplified to b(4a+3b)=0 , take a=1, b=0 or a=3, b=-4 to get the linear equation x=3 or 3x-4y+23=0.

Let this line be: y=k(x-3)+8

The equation for the circle is:

x²+(y-2)²=25

Let the distance from the center of the circle to the straight line be d, then there is:

d²=r²-(ab/2)²

i.e.: |3k-2+8|/(k²+1)=25-16|(3k+6)|=9(k²+1)

When 3k+6 0 there is:

3k+6=9k+9 i.e.: 3k -k+1=0 There is no real solution to this equation!

When 3k+6 0 there is:

3k-6=9k +9 i.e.: 3k +k-5=0 solution: k=1 3 (rounded) or k=-5

Then the linear equation is: y=-5(x-3)+8

The center of the circle (0,2), the radius = 5, through the center of the circle t to do tm perpendicular to the chord ab, from the Pythagorean theorem to know that tm = 3, then the center of the circle is (0, 2), the radius of the circle d d, the circle equation x*2 + (y-2)*2 = 9 is tangent to the straight line ab, because ab passes the point (3, 8), let ab linear equation is y-8 = k (x-3), connect these two equations, because tangent, so =0, the solution k=3 4, so the straight line ab equation is y-8 = 3 4 (x-3).

Solution: From what is known, it can be obtained: the center of the circle c is the origin, so let the linear equation be y=k(x-1)+2, and reduce it to the general equation as: y-kx+k-2=0

By the conditions: the straight line intersects ab, the distance of ab is 2 3, the radius of the circle is 2, according to the Pythagorean theorem, the center of the circle to the straight line ab

The distance is d=1The formula for the distance from a point to a straight line: (0,0) to y-kx+k-2=0 is 1, then k=3 4 can be found

The linear equation is: y=3 4*x+5 4

The center of the circle is (0,0), assuming that the linear equation is y-2=k(x-1), and the general equation is reduced to y-kx+k-2=0

By the conditions: the straight line intersects ab, the distance of ab is 2 3, the radius of the circle is 2, according to the Pythagorean theorem, the center of the circle to the straight line ab

The distance is d=1The formula for the distance from a point to a straight line: (0,0) to y-kx+k-2=0 is 1, then k=3 4 can be found

The linear equation is: y=3 4*x+5 4